So did I do this right:
1) (√3x+1) = (√16)
2) 3x+1=4 (because the √16= 4)
3) 3x+1-1=4-1
4) 3x=3
5) 3x/3=3/3
6) x= 1 (because 3/3=1)
Solve the equation (3x+1)^2=16 by the Square Root Method.
4 answers
actually, if
(3x+1)^2=16
then 3x+1 = ± 4
3x+1 = 4 or 3x+1 = -4
3x = 3 or 3x = -5
x = 1 or x = -5/3
(3x+1)^2=16
then 3x+1 = ± 4
3x+1 = 4 or 3x+1 = -4
3x = 3 or 3x = -5
x = 1 or x = -5/3
Thank you reiny
(3x+1)^2=16
(3x+1)^2 = 4^2
so
3x + 1 = 4
3x = 3
x = 1
of course also
(3x+1)^2=16
(3x+1)^2= (-4)^2 = 16 just like (+4)^2 = 16
3x+ 1 = -4
3x = -5
x = -5/3
(3x+1)^2 = 4^2
so
3x + 1 = 4
3x = 3
x = 1
of course also
(3x+1)^2=16
(3x+1)^2= (-4)^2 = 16 just like (+4)^2 = 16
3x+ 1 = -4
3x = -5
x = -5/3